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Ilia_Sergeevich [38]
3 years ago
5

A gas is allowed to expand from a volume of 400 ml to 2000 ml at a constant temperature. If the the initial pressure is 3 atm, c

alculate the final pressure exerted by the gas
Chemistry
1 answer:
Lostsunrise [7]3 years ago
8 0

Answer:

P_{2}) = 0.6 \; atm

Explanation:

<u>Given the following data;</u>

Initial volume = 400 mL

Final volume = 2000 mL

Initial pressure = 3 atm

To find the final pressure P2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

PV = K

P_{1}V_{1} = P_{2}V_{2}

Substituting into the equation, we have;

3 * 400 = P_{2}* 2000

1200 = 2000P_{2}

P_{2}) = \frac {1200}{2000}

P_{2}) = 0.6 \; atm

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Water has a density of 1.0 g/ml. which of these objects will float in water? object i: mass = 50.0 g; volume = 40.2 ml object ii
Oksana_A [137]
<span>Object I would be it</span>
8 0
3 years ago
suppose a student poured equal volumes of water and naoh in two different flasks. he forgets to label his flasks and is confused
ArbitrLikvidat [17]

The one with higher temperature is the one with NaOH as heatis given off during the neutralization reaction that occurs.

<h3>What is volume?</h3>

Volume can be defined as the amount of space a substance or an objects occupies usually in a closed container.

Volume is measures in litres.

When water is added to dilute acid like HCl, they become more dilute.

When NaOH is added to HCl, a neutralization reaction occurs.

The student will determine the contents of the flasks by adding 10 ml of hcl to each flask. If the NaOH reacts with the Hcl, there will be an increase in temperature.

The increase in temperature is due to the heat of neutralization of the reaction between NaOH and HCl.

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5 0
1 year ago
What is the [OH-] concentration if the pOH is 1.5?​
Olin [163]

Answer:

[OH-] = 10^-1.5 = 0.0316 M

Explanation:

4 0
3 years ago
calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg
alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

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6 0
2 years ago
51.7ml at 27 Celsius and 90kpa to stp
never [62]
STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K*  90^10^3Pa/ 300.15K *  1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml
6 0
3 years ago
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